Word problems about speed

Before you get into distance, time and speed word problems, take a few minutes to read this first and understand: How to build your credit score in USA as an international student.

Problems involving Time, Distance and Speed are solved based on one simple formula.

Distance = Speed * Time

Which implies →

Speed = Distance / Time   and

Time = Distance / Speed

Solution

Solution

Solution

Distance covered = 4*40 = 160 miles

Example 4. If a person walks at 4 mph, he covers a certain distance. If he walks at 9 mph, he covers 7.5 miles more. How much distance did he actually cover?

Now we can see that the direct application of our usual formula Distance = Speed * Time or its variations cannot be done in this case and we need to put in extra effort to calculate the given parameters.

Let us see how this question can be solved.

Solution

For these kinds of questions, a table like this might make it easier to solve.

Distance	Speed	Time
d	4	t
d+7.5	9	t

 
Let the distance covered by that person be ‘d’.

Walking at 4 mph and covering a distance ‘d’ is done in a time of ‘d/4’

IF he walks at 9 mph, he covers 7.5 miles more than the actual distance d, which is ‘d+7.5’.

He does this in a time of (d+7.5)/9.

Since the time is same in both the cases →

d/4 = (d+7.5)/9            →        9d = 4(d+7.5)   →        9d=4d+30        →        d = 6.

Solution

Here, we see that the distance is same.

Let us assume that its usual speed is ‘s’ and time is ‘t’, then

Distance	Speed	Time
d	s	t min
d	S+1/3	t+30 min

 
s*t = (1/3)s*(t+30)      →        t = t/3 + 10      →        t = 15.

So the actual time taken to cover the distance is 15 minutes.

Note: Note the time is expressed in terms of ‘minutes’. When we express distance in terms of miles or kilometers, time is expressed in terms of hours and has to be converted into appropriate units of measurement.

Solved Questions on Trains

Example 1. X and Y are two stations which are 320 miles apart. A train starts at a certain time from X and travels towards Y at 70 mph. After 2 hours, another train starts from Y and travels towards X at 20 mph. At what time do they meet?

Solution

Let the time after which they meet be ‘t’ hours.

Then the time travelled by second train becomes ‘t-2’.

Now,

Distance covered by first train+Distance covered by second train = 320 miles

70t+20(t-2) = 320

Solving this gives t = 4.

Solution

Let the speed of the first train be ‘s’.

Distance covered by the first train in (2+4) hours = Distance covered by second train in 4 hours

Therefore, 6s = 60*4

Solving which gives s=40.

So the slower train is moving at the rate of 40 mph.
 

Questions on Boats/Airplanes

For problems with boats and streams,

Speed of the boat upstream (against the current) = Speed of the boat in still water – speed of the stream

[As the stream obstructs the speed of the boat in still water, its speed has to be subtracted from the usual speed of the boat]

Speed of the boat downstream (along with the current) = Speed of the boat in still water + speed of the stream

[As the stream pushes the boat and makes it easier for the boat to reach the destination faster, speed of the stream has to be added]

Similarly, for airplanes travelling with/against the wind,

Speed of the plane with the wind = speed of the plane + speed of the wind

Speed of the plane against the wind = speed of the plane – speed of the wind

Let us look at some examples.

Example 1. A man travels at 3 mph in still water. If the current’s velocity is 1 mph, it takes 3 hours to row to a place and come back. How far is the place?

Solution

Let the distance be ‘d’ miles.

Time taken to cover the distance upstream + Time taken to cover the distance downstream = 3

Speed upstream = 3-1 = 2 mph

Speed downstream = 3+1 = 4 mph

Solution

Let the speed of the plane be ‘a’ and that of the wind be ‘w’.

Our table looks like this:
 

	Distance	Speed	Time
With the wind	2400	a+w	4
Against the wind	2400	a-w	6

 
4(a+w) = 2400 and 6(a-w) = 2400

Expressing one unknown variable in terms of the other makes it easier to solve, which means

a+w = 600 → w=600-a

Substituting the value of w in the second equation,

a-w = 400

a-(600-a) = 400 → a = 500

The speed of the plane is 500 kmph and that of the wind is 100 kmph.
 

More solved examples on Speed, Distance and Time

Example 1. A boy travelled by train which moved at the speed of 30 mph. He then boarded a bus which moved at the speed of 40 mph and reached his destination. The entire distance covered was 100 miles and the entire duration of the journey was 3 hours. Find the distance he travelled by bus.

Solution

	Distance	Speed	Time
Train	d	30	t
Bus	100-d	40	3-t

Let the time taken by the train be ‘t’. Then that of bus is ‘3-t’.

The entire distance covered was 100 miles

So, 30t + 40(3-t) = 100

Solving which gives t=2.

Substituting the value of t in 40(3-t), we get the distance travelled by bus is 40 miles.

Alternatively, we can add the time and equate it to 3 hours, which directly gives the distance.

d/30 + (100-d)/40 = 3

Solution

Our table looks like this.

	Distance	Speed	Time
1st part of journey	d	100	t
2nd part of journey	630-d	110	6-t

Assuming the distance covered in the 1^st part of journey to be ‘d’, the distance covered in the second half becomes ‘630-d’.

Assuming the time taken for the first part of the journey to be ‘t’, the time taken for the second half becomes ‘6-t’.

From the first equation, d=100t

The second equation is 630-d = 110(6-t).

Substituting the value of d from the first equation, we get

630-100t = 110(6-t)

Solving this gives t=3.

Solution

	Distance	Speed	Time
First person	d	3	t
Second person	20-d	4	t

 
Assuming the distance travelled by the first person to be ‘d’, the distance travelled by the second person is ’20-d’.

The time is ‘t’ for both of them because when they meet, they would have walked for the same time.

Since time is same, we can equate as

d/3 = (20-d)/4

Solving this gives d=60/7 miles (8.5 miles approximately)

Then t = 20/7 hours

So the two persons meet after 2 6/7 hours.
 

Practice Questions for you to solve

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Problem 1: Click here

Answer 1: Click here

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Problem 2: Click here

Answer 2: Click here

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The following velocity word problems will strengthen your knowledge of
speed and velocity. In the end, the difference between speed and velocity
should be clear. 

Problem #1: Two cars are traveling on US 301 south to go to the same store that is 10 miles way.

Car #1 makes it 15 minutes.

Car #2 makes it in 15 minutes.

Do the cars have the same average speed? How about velocity?

Since the cars make it at the same time, they must have the same average speed. 

15 minutes is 0.25 hour, so average speed = 10 divided by 0.25

Average speed = 40 miles per hour.

We know the cars are traveling in the same direction.

However, just knowing the average speed is not enough information to conclude that the cars have the same velocity. 

At some point, it is possible that the cars could have had the same
instantaneous speed of 30 miles per hour. In this specific case, you can
say that they had the same velocity.

Challenging velocity word problems

Problem #2: A car going westward has a cruising speed
of 50 miles per hour. Another car that is 10 miles away and
traveling eastward has a cruising speed of 50 miles per hour.

1) Do both cars have the same speed ? 2) Do they have the same
velocity ? 3) Do the cars have a constant velocity ? 4) How far will
both cars be in half an hour after they pass each other? 5) After how
long will the cars meet?

Solution :

1) The cars have the same speed since they are both traveling with a speed of 50 miles per hour.

2) The cars have different velocity since they are moving in opposite directions.

3)
Each car has a constant velocity since the speed does not
change while the car is cruising. Furthermore, the car is traveling in a straight
line.

Therefore, after half an hour both cars will be 25 miles away.

However,
recall that they were 10 miles away from each other. This means that
they will meet half way or after they have driven 5 miles. they will be
20 miles away after they pass each other. (25 — 5 = 20)

5) The formula to get the speed  is

We need to find the time, so we must rewrite the speed formula.

By the same fashion,

Since the cars are 10 miles away from each other, d = 10 miles.

However, if they maintain a constant velocity, they will meet half way or 5 miles. We know the speed is 50 miles per hour.

t = 0.10 hour.

How many minutes is 0.10 hour ?

Just multiply 0.10 by 60 and you will get 6.

So the cars will meet in 6 minutes.

Problem #3: Your turn to solve a velocity word problem!
A
car is driving on 95 south to go to New York that is 250 miles way. The
car makes it to New York in 5 hours. Did the car have a constant
velocity?

If you cannot solve this problem, try reading again the solution to the two velocity word problems above.

Problem 1 :

David drove for 3 hours at a rate of 50 miles per hour, for 2 hours at 60 miles per hour and for 5 hours at a rate of 70 miles per hour. What was his average speed for the whole journey?

Solution :

Step 1 :

Formula for average speed is

= Total distance/Total time taken

And also, formula for the distance is

= Rate ⋅ Time

Step 2 :

Distance covered in the first 3 hours is

= 50 ⋅ 3

= 150 miles

Distance covered in the next 2 hours is

= 60 ⋅ 2

= 120 miles

Distance covered in the last 5 hours is

= 70 ⋅ 5

= 350 miles

Step 3 :

Then, total distance is

= 150 + 120 + 350

= 620 miles

Total time is

= 3 + 2 + 5

= 10 hours

Step 4 :

So, the average speed is

= 620/10

= 62

So, the average speed for the whole journey is 62 miles per hour.

Problem 2 :

Jose travels from the place A to place B at a certain speed. When he comes back from place B to place A, his speed is 60 miles per hour. If the average speed for the whole journey is 72 miles per hour, find his speed when he travels from the place A to B.

Solution :

Step 1 :

Let a be the speed from place A to B.

Speed from place B to A = 60 miles/hour.

Step 2 :

Here, both the ways, he covers the same distance.

Then, formula to find average speed is

= 2xy/(x + y)

Step 3 :

x —-> speed from place A to B

x = a

y —-> speed from place B to A

y = 60

Step 4 :

Given : Average speed  is 72 miles/hour.

(2 ⋅ a ⋅ 60)/(a + 60) = 72

120a = 72(a + 60)

120a = 72a + 4320

48a = 4320

a = 90

So, the speed from place A to B is 90 miles per hour.

Problem 3 :

David travels from the place A to place B at a certain speed. When he comes back from place B to place A, he increases his speed 2 times. If the constant-speed for the whole journey is 80 miles per hour, find his speed when he travels from the place A to B.

Solution :

Step 1 :

Let a be the speed from place A to B.

Then, speed from place B to A = 2a

Step 2 :

The distance traveled in both the ways (A to B and B to A) is same.

So, the formula to find average speed is

= 2xy/(x + y)

Step 3 :

x —-> Speed from place A to B

x = a

y —-> Speed from place B to A

y = 2a

Step 4 :

Given : Average speed = 80 miles/hour

(2 ⋅ a ⋅ 2a)/(a + 2a) = 80

4a²/3a = 80

4a/3 = 80

a = 60

So, the speed from place A to B is 60 miles per hour. 

Problem 4 :

A person takes 5 hours to travel from place A to place B at the rate of 40 miles per hour. He comes back from place B to place A with 25% increased speed. Find the average speed for the whole journey.

Solution :

Step 1 :

Speed (from A to B) = 40 miles/hour

Speed (from B to A) = 50 miles/hour (25% increased)

Step 2 :

The distance traveled in both the ways (A to B and B to A) is same.

So, the formula to find average distance is

= 2xy/(x + y)

Step 3 :

x —-> speed from place A to B

x = 40

y —-> speed from place B to A

y = 50

Step 4 :

Average speed = (2 ⋅ 40 ⋅ 50)/(40 + 50)

= 44.44

So, the average speed for the whole journey is about 44.44 miles/hour.

Problem 5 :

Speed (A to B) = 20 miles/hour

Speed (B to C) = 15 miles/hour

Speed (C to D) = 30 miles/hour

If the distances from A to B, B to C and C to D are equal and it takes 3 hours to travel from A to B, find the average speed from A to D.

Solution :

Step 1 :

Formula to find distance is

= Rate ⋅ Time

Distance from A to B is

= 20 ⋅ 3

= 60 miles

Given : Distance from A to B, B to C and C to D are equal.

Total distance from A to D is

= 60 + 60 + 60

= 180  miles

Step 2 :

Formula to find time is

= Distance/Speed

Time (A to B) = 60/20 = 3 hours

Time (B to C) = 60/15 = 4 hours

Time (C to D) = 60/30 = 2 hours

Total time taken from A to D is

= 3 + 4 + 2

= 9 hours

Step 3 :

Formula to find average speed is

= Total distance/Total time

= 180/9

= 20

So, the average speed from A to D is 20 miles per hour.

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To solve the word problems on speed distance time we need to
know the relationship between speed (rate), distance and time.

  Speed = Distance
                Time

    Time = Distance
               Speed

                                           Distance = Speed × Time

The formula is used to solve different questions on distance
covered, time taken, uniform speed and variable speed problems.

The detailed explanation will help us to understand how to
solve the word problems on speed distance time.

1. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. What is the rate of the plane in still air and what is the rate of the wind?

Solution:

Let the speed of the
air plane in still air be x km/hr

And the speed of the
wind be y km/hr.

Then speed of the
airplane going with the wind = (x + y) km/hr

and speed of the
airplane going against the wind = (x — y) km/hr.

We know that,

                                 Distance = Speed
× Time

                     or,     Speed = Distance
                                                Time

According to the
problem,

An airplane travels 4362 km against the wind in 6 hours

     x — y = 4362/6

or, x — y = 727 ———— (1)

Again, the
airplane travels 5322 km with the wind in the same amount of time i.e. 6 hours

     x + y = 5322/6

or, x + y = 887 ————
(2)

Now add (1) and (2) we get,

 x — y =  727

 x + y =  887

2x     = 1614

or, 2x/2 = 1614/2,
(Divide both sides by 2)

or, x = 807

Now substitute the
value of value of x = 807 in equation (2) we get,

  807 + y = 887

-807        -807, (Subtract 407 from both sides)

          y = 80

Answer: Rate
of the plane in still air = 807 km/hr

            Rate
of the wind = 80km/hr

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